$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$ How to evaluate $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})$$
I'm completely stuck into it.
 A: Hint: $\sqrt{n^2+n}-\sqrt{n^2+1}=\frac{n-1}{\sqrt{n^2+n}+\sqrt{n^2+1}}$
A: We can use the approximation 
$$(1+x)^\alpha\sim_0 1+\alpha x$$
then in our case we have $\alpha=\frac{1}{2}$  so
\begin{align}\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt{n^2+1})&=\lim_{n\to\infty}n(\sqrt{1+\frac{1}{n}}-\sqrt{1+\frac{1}{n^2}})\\&=\lim_{n\to\infty}n(1+\frac{1}{2n}-1-\frac{1}{2n^2})=\frac{1}{2}.\end{align}
A: Multiply and divide by the quantity $\sqrt{n^{2}+n} + \sqrt{n^{2}+1}$
A: A useful general approach to limits is, in your scratch work, to take every complicated term and replace it with a similar approximate term.
As $n$ grows large, $\sqrt{n^2 + n}$ looks like $\sqrt{n^2} = n$. More precisely,
$$ \sqrt{n^2 + n} = n + o(n) $$
where I've used little-o notation. In terms of limits, this means
$$ \lim_{n \to \infty} \frac{\sqrt{n^2 + n} - n}{n} = 0 $$
but little-o notation makes it much easier to express the intuitive idea being used. 
Unfortunately, $\sqrt{n^2 + 1}$ also looks like $n$. Combining these estimates,
$$ \sqrt{n^2 + n} - \sqrt{n^2 + 1} = (n + o(n)) - (n + o(n)) = o(n) $$
Unfortunately, this cancellation has clobbered all of the precision of our estimates! All this analysis reveals is
$$ \lim_{n \to \infty} \frac{\sqrt{n^2 + n} - \sqrt{n^2+1}}{n} = 0 $$
which isn't good enough to answer the problem. So, we need a better estimate.

A standard way to get better estimates is differential approximation. While the situation at hand is a little awkward, there is fortunately a standard trick to deal with square roots, or any power:
$$ \sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} $$
and now we can invoke differential approximation (or Taylor series)
$$ f(x+h) = f(x) + h f'(x) + o(h) $$
with $f(x) = 1 + \frac{1}{x}$ at $x=1$ to get
$$ \sqrt{n^2 + n} = n \left( 1 + \frac{1}{2n} + o\left(\frac{1}{n} \right)\right)
= n + \frac{1}{2} + o(1) $$
or equivalently in limit terms,
$$ \lim_{n \to \infty} \sqrt{n^2 + n} - n - \frac{1}{2} = 0$$
similarly,
$$ \sqrt{n^2 + 1} = n + o(1)$$
and we get
$$ \lim_{n \to \infty} \sqrt{n^2 + n} - \sqrt{n^2 + 1}
=  \lim_{n \to \infty} (n + \frac{1}{2} + o(1)) - (n + o(1))
=  \lim_{n \to \infty} \frac{1}{2} + o(1) = \frac{1}{2} $$

If we didn't realize that trick, there are a few other tricks to do, but there is actually a straightforward way to proceed too. Initially, simply taking the Taylor series for $g(x) = \sqrt{n^2 + x}$ around $x=0$  doesn't help, because that gives
$$ g(x) = n + \frac{1}{2} \frac{x}{n} + o(x^2) $$
the Taylor series for $h(x) = \sqrt{x^2 + x}$ doesn't help either. But this is why we pay attention to the remainder term! One form of the Taylor remainder says that:
$$ g(x) = n + \frac{1}{2} \frac{x}{n} - \frac{1}{8} \left( n^2 + c \right)^{-3/2} x^2$$
for some $c$ between $0$ and $x$. It's easy to bound this error term for $x > 0$.
$$ \left| \frac{1}{8} \left( n^2 + c \right)^{-3/2} x^2 \right|
\leq  \left| \frac{1}{8} \left( n^2 \right)^{-3/2} x^2 \right|
= \left| \frac{x^2}{8 n^3} \right| $$
So, for $x > 0$,
$$ g(x) = n + \frac{1}{2} + O\left( \frac{x^2}{n^3} \right) $$
(note I've switched to big-O). Plugging in $n$ gives
$$ g(n) = n + \frac{1}{2} + O\left( \frac{1}{n} \right) $$
which gives the approximation we need (better than we need, actually). (One could, of course, simply stick to limits rather than use big-O notation)
This is not the simplest way to solve the problem, but I wanted to demonstrate a straightforward application of the tools you have learned (or will soon learn) to solve a problem in the case that you can't find a 'clever' approach.
